11. Lemma 2.1.1- Cauchy Integral Theorem for rectangles

Let $f : U \to C$ be a holomorphic function, and let $R = [a, b] \times [c, d]$ be a rectangle contained in $U$ . Then for any parametrization of $δ R$ , say $γ : [t_{0}, t_{1}] \to U$ , we have:
:?:

\oint_{γ} f (z) d z := \int_{t_{0}}^{t_{1}} f (γ (t)) γ^{'} (t) d t = 0

PROOF:

Split up $R$ into 4 equal sized smaller rectangles. Then our integral is the sum of the integrals over these smaller rectangles (the sides inside cancel). Pick $Q^{(1)}$ to be the rectangle with the greatest absolute value of the integral. Then we have:

| \oint_{R} f (z) d z | \leq 4 | \oint_{Q^{(1)}} f (z) d z |

Repeat this process many times so that we have:

3. N o t i c e t h a t t h e s e r e c t a n g l e s a r e n e s t e d, a n d t h e i r d i a m e t e r s g o t o z e r o . I f w e p i c k $ a_{n} $ t o b e t h e c e n t e r o f $ Q^{(n)} $, t h e n $ a_{n} $ c o n v e r g e s t o s o m e $ z_{0} \in R $ . $ f $ i s c o m p l e x d i f f e r e n t i a b l e h e r e . C o m e u p w i t h t h e e s t i m a t e $ f (z) = f (z_{0}) + f^{'} (z_{0}) (z - z_{0}) + χ (z) $, w i t h $ lim_{z \to z_{0}} \frac{| χ (z) |}{| z - z_{0} |} = 0 $ . 4. U s e t h e e q u a t i o n a b o v e t o s p l i t u p $ | \oint_{Q^{(n)}} f (z) d z | $ a n d u s e t h a t $ f (z_{0}) $ a n d $ f^{'} (z_{0}) (z - z_{0}) $ h a v e a n t i d e r i v a t i v e s t o g e t :

|\oint_{R}f(z)dz| \leq 4^n |\oint_{Q^{(n)}}\chi(z)dz|

5. B o u n d t h i s e q u a t i o n u s i n g t h e l i m i t f r o m s t e p 3. T h e n s i m p l i f y t h i s u n t i l w e h a v e s o m e t h i n g t h a t l o o k s l i k e :

|\oint_{R}f(z)dz| \leq \epsilon M